关于android手机图案锁屏的计数

无心人

2=>1,3,4,5,6,7,9 3=>2,4,5,6,8 4=>1,2,3,5,7,8,9 5=>1,2,3,4,6,7,8,9

无心人

6=>1,2,3,5,7,8,9 7=>2,4,5,6,8 8=>1,3,4,5,6,7,9 9=>2,4,5,6,8

wayne

不错，这种方法好是好，只是有一种情况没考虑吧。 比如，前面如果已经经过5的话，后面4可以直接跳到6的 参考6#可行方案的2516图 路径{8,5,7,4,6}

无心人

待我写个程序来

无心人

2. #include <stdio.h>
4. int mask[10] = {1,0,0,0,0,0,0,0,0,0};
5. int next[10][10] =
6. {0,0,0,0,0,0,0,0,0,0,
7. 5,2,4,5,6,8,0,0,0,0,
8. 7,1,3,4,5,6,7,9,0,0,
9. 5,2,4,5,6,8,0,0,0,0,
10. 7,1,2,3,5,7,8,9,0,0,
11. 8,1,2,3,4,6,7,8,9,0,
12. 7,1,2,3,5,7,8,9,0,0,
13. 5,2,4,5,6,8,0,0,0,0,
14. 7,1,3,4,5,6,7,9,0,0,
15. 5,2,4,5,6,8,0,0,0,0,
16. };
17. int counter[10] = {0,0,0,0,0,0,0,0,0,0};
18. int graph[10] = {0,0,0,0,0,0,0,0,0,0};
19. int nextNumber = 0;
20. int graphPostion = 0;
22. void first( void )
23. {
24. int i;
25. for (i = 1; i <= 9 ; i ++)
26. {
27. mask[i] = 1;
28. graph[graphPostion] = i;
29. circle( graphPostion + 1 );
30. mask[i] = 0;
31. }
32. }
34. void circle( int graphPostion )
35. {
36. int i;
37. int noPostion = 1;
38. int n = graph[graphPostion - 1];
39. for (i = 1; i <= next[n][0]; i ++)
40. {
41. if (mask[next[n][i]] == 0)
42. {
43. noPostion = 0;
44. mask[next[n][i]] = 1;
45. graph[graphPostion] = next[n][i];
46. counter[graphPostion + 1] ++;
47. circle( graphPostion + 1 );
48. mask[next[n][i]] = 0;
49. }
50. }
52. if (noPostion)
53. {
54. printf("find: ");
55. for (i = 0; i < graphPostion; i ++)
56. {
57. printf("%1d ", graph[i]);
58. }
59. printf("\n");
60. }
61. }
63. int main ( void )
64. {
65. int i;
66. first( );
67. for (i = 4; i <= 9; i ++)
68. printf("%d: %d\n", i, counter[i]);
69. return 0;
70. }

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4: 1400 5: 5328 6: 16032 7: 35328 8: 49536 9: 32256

wayne

1. PointsToBeChecked[{{a_,b_},{c_,d_}}]:=Module[{g=GCD[c-a,d-b]},If[g>1,{a+# (c-a)/g,b+# (d-b)/g}&/@Range[g-1],{}]];
2. n=3;ddd=Flatten[Table[{{i,j}},{i,0,n-1},{j,0,n-1}],1];
3. Table[{ii,Product[n^2-k,{k,0,ii-1}],ddd=Select[Flatten[Table[Join[ii,List@#]&/@Complement[Flatten[Array[List,{n,n}]-1,1],ii],{ii,ddd}],1],Length[PointsToBeChecked[#[[-2;;-1]]]]==0&];Length[ddd]},{ii,2,n^2}]

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对算法做了改进，反而跟无心人的答案对上了。 {2, 72, 56}, {3, 504, 304}, {4, 3024, 1400}, {5, 15120, 5328}, {6, 60480, 16032}, {7, 181440, 35328}, {8, 362880, 49536}, {9, 362880, 32256}}

wayne

算得4*4的情况： 2,240,172 3,3360,1696 4,43680,15580 5,524160,132264 6,5765760,1029232 7,57657600,7286016 后面就算不动了

无心人

证明俺算法没错了哦

mathematica

取出所有的排列,然后一个一个地判定!

gogdizzy

看了下无心人的程序，能计算#6给出的合法情况吗？ 5->2->7->3->4 因为5是已经标记过的，所以可以从7直接到3了。